An Experiment Question

An Experiment Question PhL 11.25.15 1.15.16 This remains a tough issue for me, not sure how related to wedge doc. Definitely keep this doc around because it may become important. Consider: Experiment A : (active) va → v'a = Rarvr (ei)a → (ei)a Mab → M'ab = RarRbsMrs apparatus changes basis (axes) stay the same Question 1: What happens to the metric tensor in Experiment A? Is it associated with the apparatus (in which case it should change), or is it associated with the basis (in which case it should not change). Before tackling that question, here are some others to deal with first: Question 2: Is the above Experiment description meaningful when the transformation is something other than a rotation? (a) If it is a rotation (with g = g' = 1), I say v' = Rv and the components of v' are vi → v'i = v' ei ≠ vi not vi → v'i = v' e'i = v ei = vi This is what is meant by the above description of the experiment: v (part of apparatus) changes but the basis vectors ei do not change. (b) Suppose the transformation is a stretch thing where each of the three axes is stretched differently. We can then speak about a "warped" apparatus which is in x-space. There is some R such that v' = Rv where R is the stretch matrix. The apparatus is some kind of machine which has parts which have velocities. This warped machine sits in x-space where we, also in x-space, look at it. We want to measure the velocities of the parts of the warped machine. Note" We are in physical space where gij = δij.The metric tensor g'ij = RiaRjb gab = RiaRja is something different from δi,j. We do as above for this experiment, same as for a rotation, v'i = v' ei ≠ vi So I think the Experiment A idea would apply just as well to this stretching case as to the previous rotation case. (c) The general case is just a combined rotation and stretching. Answer: Yes, Experiment A's description applies when R is not just a rotation. Now look again at the stretch situation. We can say v'i = v' ei = gab(v')a(ei)b = δab(v')a(ei)b = (v')a(ei)a = (v')a(ei)a In doing this dot product, since we are in our physical space and since that is where we look at vectors which are part of the warped machine and where we measure components of vectors, we use the physical space metric tensor. But we still used the transformed vector v' along with the untransformed basis vector ei. This leads me to conjecture: In Experiment A, you measure things with the untransformed ei and with the untransformed gij. You think then of gij as belonging to the space which has basis ei. This is emphasized when you write gij = ei ej. But just saying gij = ei ej does not clinch the argument for me because it is circular in that the dot product involves g. Suppose the physical space is also stretched for some reason so that gij ≠ δij, and this is the space we work in for some reason. I would say that gij = ei ej = gab(ei)a(ej)b = gabδiaδjb = gij so at least things are consistent. So even in this case I would say that gij → gij as part of Experiment A. So I rewrite the above as Experiment A : (active) va → v'a = Rarvr (ei)a → (ei)a Mab → M'ab = RarRbsMrs gab → gab apparatus changes basis (axes) stay the same Anything involving just the basis vectors, perhaps the affine connection, would be unchanged. Another thing would be eiej -- it would remain unchanged. So back to the original question: Question: What happens to the metric tensor in Experiment A? Is it associated with the apparatus (in which case it should change), or is it associated with the basis (in which case it should not change). Answer: It is associated with the basis vectors and does not change, so gij → gij under this Experiment A transformation. _______________________________________________________________________- New Question Question 3: In the curvilinear coordinates development of tensor doc, what corresponds to the machine which is getting warped in Experiment A above? The machine is a skewed n-piped which happens to align with the ei vectors. We do the experiment A transformation and this n-piped becomes a perfect cube in x-space, but the ei vectors stay where they were, so they do not align with the cube edges. The n-piped edges transform as q' = Rq and that is why we get the cube, because q' is like e'n. We measure the edges of this perfect cube using the en basis vectors in x-space and in that way we find the components of the cube edges. If q' = Reiis a warped edge, then q'k = q' ek = a coordinate of the warped machine's edge = (Rei) ek = gab (Rei)a(ek)b = (Rei)a(ek)a = Rac (ei)c(ek)a = RacRicRka = δajRka = Rkj _______________________________________________________________________- New Question Question: Discuss the transformation of a tensor expansion in various Experimental contexts: The starting position is this T = Σii....i T'ii....i (ei ei ..... ei) . V = Σn V'n en or T = Σii....i T(e)ii....i (ei ei ..... ei) . V = Σn V(e)n en We have not transformed anything yet! There is however an underlying R in that en = Rmn un which R relates en to the axis-aligned un. Experiment A. Now what happens if we do an Experiment A active transformation R' on this "machine" ? We do this with a transformation called R', to avoid confusion with the "underlying R" mentioned above. Then, T' = Σii....i [T(e)] ' ii....i (ei ei ..... ei) . V = Σn [V(e)]' n en where [T(e)] ' ii....i = R'ijR'ij...R'ij [T(e)]jj....j [V(e)]' n = R'nm [V(e)]m Then T' is some new tensor in Vk that we got by doing an active transformation on tensor T. Since this notation is quite cumbersome, though accurate, we now make some changes: 1. Ignore the underlying R in en = Rmn un . Or rename it to be en = Rmn un . 2. Rename the R' transformation of Experiment A to be R. 3. Suppress the (e) labels on tensors, so T(e) → T. This conflicts with our Section 2.6 convention that the default tensor type was T = T(u). We are thus hereby changing that default so unmarked tensors are all (e) type tensors associated with (e) space. Then we rewrite the equations above: Before the active transformation: T = Σii....i Tii....i (ei ei ..... ei) . V = Σn Vn en After an active transformation by R we find a new tensor T' in Vk which is T' = Σii....i T'ii....i (ei ei ..... ei) . V = Σn V'n en where T' ii....i = RijRij...Rij Tjj....j V' n = Rnm Vm I think this makes things be compatible with the notation of Chapter 5, such as the general expansion shown in (5.2.1). So this is the "missing link" in the transition from Chapter 2 to Chapter 5. I think it accounts for much of my confusion! I think all of the above is OK, but below is more conjectural stuff that needs work. Example. Look at how this functional transforms under Experiment A xform R: λi(v) → λi'(v) = ei v' = λi(v') = (v')i = Rijvj = Rijλj(v) so our transformation rule is λi(v) → λi'(v) = Rijλj(v) or λi → λi' = Rijλj so λi transforms as a rank-1 tensor (vector), as expected, in an Experiment A transformation. What happens to the functional α(v) ? α(v) = Σiαiλi(v) → α'(v) = Σiαiλ'i(v) = ΣiαiRijλj(v) = ΣiαiRijvj = Σiαiv'i = Σiαiλi(v') = α(v') so the rule is α'(v) = α(v') under this Exp A transformation Not so clear that it is a rank-1 tensor here? I would guess that T'(v1,v2) = T(v'1,v'2) but this differs from which I have in (D.4.6), although it is close ! Have to clean this up. Needs a better description of the experiments: Experiment A : (active transformation) tensors: T = ΣITIeI → T' = ΣIT'IeI T → T' |T> = ΣI<eI|T> |eI> |T'> = ΣI<eI|T'>|eI> tensor components: TI → T'I = RIJTJ <eI| T> → <eI| T'> = RIJ<eJ| T> basis vectors: eI → eI |eI> → |eI> Experiment B : (passive transformation) tensors: T = ΣITIeI → T = ΣIT'Ie'I |T> = ΣI<eI|T> |eI> → |T> = ΣI<eI|T'> |e'I> <eI|T'> = <e'I|T> ?? tensor components: TI → TI' = RIJTJ <eI| T> → <eI| T'> = RIJ<eJ| T> basis vectors: eI → e'I = (R-1)IJ eJ |eI> → |e'I> = (R-1)IJ |eJ> Check: <eI|T'> = These are the only two one really cares about I think. Example: T = Σii....i Tii....i (ei ei ..... ei) . V = Σn Vn en Then T = Σii....i T'ii....i (e'i e'i ..... e'i) . V = Σn V'n e'n The key point is that the primes on the basis vectors have a different meaning. What happens for functionals? Side Question: If f(v) is linear, can you say f(Rv) = Rf(v) ? Answer: If f(v) is linear, we know that f(αv) = αf(v). But the above does not have this form because R is not a scalar. So the answer is: NO. Side Question: If v' = Rv, can you say |v'> = |Rv> = R|v> ? By same argument, NO. I used to say |Rv> = R |v> where R is the Hilbert Space operator, while R is the Rn operator. In this case <Rv| = <v| RT ?? but R is always unitary ??? ..... I am not sure I want to get into all of this old stuff right now. Example. Look at how this functional transforms under Experiment B xform R: Remember that the functional λi "has no components" the way ei does. λi(v) → λi'(v') = e'i v' = ei v = λi(v) = vi = (R-1)ijv'j = (R-1)ijλj(v') λi'(v') = (R-1)ijλj(v') λi' = (R-1)ijλj α(v) = Σiαiλi(v) α(v) → α'(v') = Σiα'iλi(v') = Σiα'iλi(v') α(v) = Σiαiλi(v) → α'(v) = Σiαiλ'i(v) = ΣiαiRijλj(v) = ΣiαiRijvj = Σiαiv'i = Σiαiλi(v') = α(v') I am wobbly on this. I have located a 2008 doc I wrote called "confusion about rotation operators" and it is exactly on this topic of the various kinds of experiments in terms of the bra-ket notation. I suspect it can shed some light on my present attempt to define these experiments. I will to study this thing since it is directly on the topics of Hilbert Space bras and kets and Experiments. I think it only deals with rotations, but I can probably generalize it.